Is there another simpler method to solve this elementary school math problem?

Question

I am teaching an elementary student. He has a homework as follows.

There are $16$ students who use either bicycles or tricycles. The total number of wheels is $38$. Find the number of students using bicycles.

I have $3$ solutions as follows.

Using a single variable.

Let $x$ be the number of students in question. The number of students using tricycles is $16-x$. The total number of wheels is the sum of the total number of bicycles times $2$ and the total number of tricycles times $3$.

$$ 2\times x + 3 \times (16-x) = 38 $$

The solution is $x=10$.

Using $2$ variables.

Let $x$ and $y$ be the number of students using bicycles and tricycles, respectively. It implies that

\begin{align} x+y&=16\\ 2x+3y&=38 \end{align}

The solution is $x=10$ and $y=6$.

Using multiples

The multiples of $2$ are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22,24, 26,28,30,32,\dotsc$

The multiples of $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,\dotsc$

The possible wheel combinations with format $(\#\text{bicycle wheels}, \#\text{tricycle wheels})$:

$\hspace{6cm} (32,6)$ but there are $18$ students

$\hspace{6cm} (26,12)$ but there are $17$ students

$\hspace{6cm} (20,18)$ there are $16$ students

$\hspace{6cm} (14, 24)$ there are $15$ students

$\hspace{6cm} (8, 30)$ there are $14$ students

$\hspace{6cm} (2,36)$ there are $13$ students

Thus the correct combination is $10$ bicycles and $6$ tricycles.

My question

Is there any other simpler method?

Answer 1

If everybody had bicycles, there would be $16 \times 2 = 32$ wheels. There are actually $38$, so that's $6$ additional wheels. Each one must be on a different tricycle. So there are $6$ tricycles, and the other $10$ students have bicycles.

Answer 2

Based on Robert Israel's method:

If everybody had tricycles, there would be $16×3=48$ wheels. There are actually 38, so that's 10 fewer wheels. Each one must be on a different bicycle. "So there are 10 bicycles (and the other 6 students have tricycles)". The parenthesized part is optional, since that was not asked for.

Answer 3

This is probably not the easiest for mental calculation, but it works.

The average number of wheels per student is $\frac{38}{16}=2+\frac38=\alpha$. So with $b,t$ the numbers of (bi/tri)cycles one has $b:t=(3-\alpha):(\alpha-2)=5:3$, and the number of bicycles is $\frac b{b+t}\times16=\frac 58\times16=10$.

Answer 4

If all students were using just a wheel, there would be 16 wheels.

If all students were using two wheels, they would be on bicycles, and there would be 32 wheels.

We have 38 wheels, so some students are using 3 wheels. 38 -32 = 6, so 6 students are using 3 wheels, and the remaining (16 - 6) will be using 2 wheels.

Answer 5

I think the "simplest" (meaning least thinking, not least work) is to create a table of all possibilities.

bike trike wheels
 0    16     48
 1    15     47
 2    14     46
 3    13     45
 4    12     44
 5    11     43
 6    10     42
 7     9     41
 8     8     40
 9     7     39
10     6     38  <- answer
11     5     37
12     4     36
13     3     35
14     2     34
15     1     33
16     0     32

In fact, once you start filling out the table, you notice a pattern. The last column decreases by one on each line. That teaches the kid that there are hidden patterns, and if they discover the pattern after a few lines, then can visually extrapolate how many lines they'll need to calculate until they get close to the answer.

It motives the concept of a linear progression ( -= 1 each time), and this begins to look like a downward slope of "something", which might be something that looks like a hill which maps to the abstract concept of a line graph with negative slope.

Also, the problem could be changed to something like 160 people, and 380 wheels and the table begins to get unwieldy. It is still the same concept, but at this point, the student will think "there must be a better way" which leads to using graphs and algebra to solve the same problem.

Answer 6

This is a little silly, perhaps, but here goes.

Since $38$ is an even number and since tricycles have an odd number of wheels (while bicycles have an even number), the number of tricycles must be even. Let's try bisection. If everyone rides a tricycle, you have $16\times3=48$ wheels, which is too many, while if no one does you have only $16\times2=32$, which is too small. So let's try $8$ and $8$, which gives $8\times3+8\times2=24+16=40$, which is too many again, so now try $4$ and $12$, giving $4\times3+12\times2=12+24=36$. That's too small. So we now know the number of tricycles must be an even number between $4$ and $8$. So it must be $6$, and the number of bicycles must be $10$.

Added 8/29/13: I can't resist adding a second solution which should also not be taken seriously as pedagogy for an elementary student but is, I think, worth pondering as a nice little exercise in logical analysis.

Make as many pairs of bicyclists and tricyclists as you can, leaving some number of unpaired cyclists. At this point we don't know whether the unpaired cyclists are on bikes or trikes, but, since the total number of cyclists ($16$) is an even number, we can say there is an even number of unpaired cyclists, who hence account for an even number of wheels. Now because the total number of wheels ($38$) is even, we can conclude there must be an even number of bike-trike pairs, since each pair accounts for an odd number ($2+3=5$) of wheels. This means the total number of paired cyclists is a multiple of $4$, which, since $16$ is also a multiple of $4$, means the number of unpaired cyclists is a multiple of $4$, so that the number of wheels they account for is a multiple of either $8$ (if they're on bikes) or $12$ (if they're on trikes).

Now let's get serious. Because there's an even number of paired cyclists, the number of wheels they account for is a multiple of $10$, ie. $10$, $20$, or $30$, leaving $28$, $18$, or $8$ wheels, respectively, for the unpaired cyclists. Neither $28$ nor $18$ is a multiple of either $8$ or $12$, leaving $8$ as the only possbility, which can only come from $4$ unpaired bicyclists and $6$ bike-trike pairs, for a total of $10$ bicyclists.

Whew!

Answer 7

Suppose that we remove $2$ wheels from each vehicle. There were $38$ wheels originally, and we removed $16\times 2=32$ wheel. This leaves us with $38-32=6$ wheels. But since all Bicycles and Tricycles had $2$ or $3$ wheels, now they must have either $0$ or $1$ wheel. Hence, there must be 6 tricycles, each with one wheel now. Put the wheels back now, there are 10 bicycles.

Alternatively:

Remove $3$ wheels from each vehicle instead of $2$, this leaves us with $38-48=-10$ wheels, which are missing on the 10 bikes.

Answer 8

Marc's idea for dummies: $38/16=2\frac 38$. So we are $3/8$ of the way from all bicycles to all tricycles (the general idea of linearity should be known if one wants to follow this approach). $\frac 38\cdot 16=6$ tricycles.

Answer 9

Each student's cycle has at least two wheels. If we account for that, 38 - 2 * 16 = 6 students have a cycle with an additional wheel.

Answer 10

Don't discount guess-and-check.

Guess-and-check can be a valid way to solve problems like this because:

• If he guesses right the first time, he will have solved the problem much faster than he would otherwise.
• It will develop arithmetic skills.

• Guess-and-check, i.e., brute force, generally suffers from being terribly inefficient. I'd argue that this will force him to do one of two things:

  1. Develop an intuition about the problem. In this case, he might notice the pattern mentioned by Mark Lakata about a linear progression. Beyond that he might think of doing a bisection, which all of a sudden becomes efficient and ceases to be brute-force. Or he might just solve the problem the algebraic way without even realizing it.

  2. He realizes the value in solving the problem mathematically.

  Either way, a lesson is learned.

Answer 11

Take variable $x$ for bicycles and $(16-x)$ for tricycles.

Now, consider the equation:

$$ 2x+3(16-x)=38$$

On solving this, we get $x=10.$ So, $10$ bicycles and 6 tricycles.

Answer 12

Robert Israel's solution is (I believe) best suited towards an elementary school student if it is graphical:

1. Sketch out 16 bicycles. Count the wheels. There are 32 wheels:

2. Start adding wheels to make tricycles until you have 38 wheels (i.e., every time you add a wheel, count up from 32). Count the number of tricycles and bicycles. Now you know there are six trikes and ten bikes:

Answer 13

I have a better way to explain this to a elementary student. XD

Like the problem, I'll say there are 16 monsters, several of them are amphisbaenas (a snake have two heads) and the remains are cerberus (a dog have three heads). They all raise their heads and I'm the master.

After my first whistle, all monsters lower one head, that's of course 16 heads. And after my seconde whistle, all monsters lower one more head, that's 32 heads in total now.

So only the 6 cerberus still raise one head. And that's the answer.

Answer 14

Draw a linear graph with $x$ the number of trike riders and $y$ the number of wheels. Then just check the $x$ value at $y=38$.

Answer 15

16 (students) X 3 (tricycle wheels) - 38 (total number of wheels) = 10 (bicycles).

10 bicycles X 2 wheels = 20 wheels 6 tricycles x 3 wheels = 18 wheels

Number of students multiplied by the number of tricycle wheels minus the total number of wheels equals the number of bicycles.

:P

Answer 16

Consider the total wheel function $f$ of two variables $x$ the number of bicyclists and $y$ the number of "tricyclists". Since every time we add a bicyclist we increase the number of wheels by two, we have $$ f_x = 2. $$ Similarly for case of tricylists, $$ f_y = 3. $$ So we have the solutions $$ f(x,y) = 2x + A(y) \\ f(x,y) = 3y + B(x) $$ which give $$ f(x,y) = 2x + 3y + C. $$ However, $f(0, 0) = 0$ as we have no wheels when we have no riders, so $C = 0$.

Since the total number of cyclists is $16$, we are only interest in the portion of $f$ that lies on the line $y = 16 - x$, that is $f(x, 16 - x)$. Morever, we what to find when the total number of wheels on this line is $38$, so $$ \begin{aligned}f(x, 16 - x) & = 38 \\ 2x + 3(16-x) & = 38 \\ -x + 48 & = 38 \\ -x & = -10 \\ x & = 10. \end{aligned} $$ Thus the solution is the point $(x,y) = (10, 6)$.

Answer 17

Simplest

Every kid has at least two wheels, $16 \times 2 = 32$ in total. The remaining $38 - 32 = 6$ wheels belong to tricycles, so there are $16 - 6 = 10$ kids on bicycles.

Trial-and-error

Choose a random integer $x$ between $0$ and $16$ repeatedly until $2x + 3(16 - x)$ equals $38$. If the equality holds, $x$ is the solution. This can be done by hand, using dice or a random number table, but the easiest way is to write a computer program (see the code here).

Root finding

The number of kids on bicycles, $x$, is the root of the first order polynomial $f(x) = 2x + 3(16 - x) - 38 = 10 - x$. The latter form shows directly that the root is $10$.

It's worth deriving the general formula. Let $m$ and $n$ be the number kids and the number of wheels, respectively. With this notation, $f(x) = 2x + 3(m - x) - n = 3m - n - x$, that is, the root $x = 3m - n$.

Answer 18

We can answer that recursively ,for example: If we have an even number we will subtract the number by $2$ and if the number is even we will subtract him by $3$,basically that's how we think .

This way maybe isn't so good to solve all the question but it's good way to explain the situation intuitively ,and if you really want to see the recursive function I'll show you.

The recursive function:

So as I said before if we have bicycle we will subtract by $2$ and if we have tricycle we will subtract by $3$ , and from that we get our recursive function:

$n$ represent the number of cycles and $k$ the number of kids.

$f(n,k)=f(n-2,k-1)+f(n-3,k-1)$

Base cases:

$$f(1,3) =1\\ f(1,2) =1$$

For each other value we will get zero.

The answer will give us the number of ways,but this was a little demonstration of our mind solve problem of this kind.